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In April 1974, Steve Prefontaine completed a 10.0 km10.0 km race in a time of 27 min27 min , 43.6 s43.6 s . Suppose "Pre" was at the 7.85 km7.85 km mark at a time of 25.0 min25.0 min . If he accelerated for 60 s60 s and then maintained his increased speed for the remainder of the race, calculate his acceleration over the 60 s60 s interval. Assume his instantaneous speed at the 7.85 km7.85 km mark was the same as his overall average speed up to that time.

Respuesta :

Answer:

a = 0.161 [tex]$m/s^2$[/tex]

Explanation:

Given :

[tex]$ d_{total}[/tex] = 10 km = 10000 m

[tex]$t_{total} $[/tex] = 27 min 43.6 s

        = 1663.6 s

[tex]$d_1$[/tex] = 7.85 km = 7850 m

[tex]$t_1$[/tex] = 25 min = 1500 s

[tex]$t_2$[/tex] = 60 s

Now the initial speed for the distance of 7.85 km is

[tex]$ v_1 = \frac{d_1}{t_1} = \frac{7850}{1500}$[/tex]  = 5.23 m/s

The velocity after 60 s after the distance of 7.85 kn is

[tex]$v_2 = v_1 + at_2$[/tex]

    = 5.23 + a(60)

The distance traveled for 60 s after the distance of 7.85 km is

[tex]$d_2 = v_1t_2+\frac{1}{2}at_2^2$[/tex]

[tex]$d_2 = (5.23)(60)+\frac{1}{2}a(60)^2$[/tex]

    = 313.8 + a(1800)

The time taken for the last journey where the speed is again uniform is

[tex]$t_3 = t_{total}-t_1-t_2 $[/tex]

   = 1663.6 - 1500 - 60

   = 103.6 s

Therefore, the distance traveled for the time [tex]$t_3$[/tex] is

[tex]$ d_3 = v_2 t_3$[/tex]

    = (5.23+60a)(103.6)

    = 541.8 + 6216 a

The total distance traveled,

[tex]$ d_{total}= d_1 + d_2 + d_3$[/tex]

Now substituting the values in the above equation for the acceleration a is

10000 = 7850 + (313.6 + 1800a) + (541.8 + 6216a)

10000 = 8706.5 + 8016a

1294.4 = 8016a

a = 0.161 [tex]$m/s^2$[/tex]

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